The forum discussion centers on the incorrect application of voltage division in AC circuit analysis, specifically regarding the calculation of current I10. The user attempted to calculate Vx using the formula Vx = 100 * (j5)/(4+j5), resulting in Vx = 60.98 + j48.78 V. However, the calculation failed because it did not account for the impedance of the 10Ω resistor and the -5j capacitor branch connected at the Vx node. The suggestion to use nodal analysis was provided as a more accurate method to determine Vx.
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Your voltage division isn't taking into account the impedance of the 10Ω and -5j capacitor branch that's also connected at the Vx node. You might try applying nodal analysis to find Vx...Ammar w said:Homework Statement
Hi,
We have this cut. :
and in the image below, I10 is calculated by using current division :
The Attempt at a Solution
I tried to find it using voltage division but the answer is wrong :
Vx(of the upper node) = 100 * [tex]\frac{j5}{4+j5}[/tex] = 60.98 + j48.78 V
=> I10 = [tex]\frac{Vx}{10 - j5}[/tex] = 2.93 + 6.34 = 6.98 [itex]\ 65.2[/itex]
which is wrong, WHY?